例6,1设一单位反馈控制系统的开环传递函数为: 6()mx)截止频率o≥44(rads),相角裕度y”≥45幅值裕 要求系统斜坡输入下的位置输出稳态误差e<0 度h≥10db,试设计系统的串联超前校正装置。 解:由e=1/k=1/<0,,有K≥10;o=√1×10=3.14,h=∞ y=180-90- aretha2=179° O=O.=4.4 -2odB/d olga=I L(a)=20kg10-20g44-20g√+4 TTTT 2.2 :10ga=-C(a)=6∴a=4 T =0.114 L() 40dB/dec n G()=1+0.456 +0.114s 10(1+0.456s 0. 100 1000 G()= s(1+s)(1+0.114s) arcsin a+1 O)=1800-90°-rcgo y=gm+y(a2)=369+1280=497°>45,h=• 设一单位反馈控制系统的开环传递函数为: 要求系统斜坡输入下的位置输出稳态误差ess≤0.1, 截止频率ωc ‘’≥4.4(rad/s),相角裕度γ‘’≥45o,幅值裕 度h‘’≥10db,试设计系统的串联超前校正装置。 例6.1 ( 1) ( ) + = s s K G s 1 10 3.14 , ' = = c o c o o 180 90 arctg 17.9 ' = − − = 4.4 '' m =c = ( ) 20lg10 20lg 4.4 20lg 1 4.4 6 ' '' 2 L c = − − + = − 0.114 1 = = a T m s s G s c 1 0.114 1 0.456 4 ( ) + + = (1 )(1 0.114 ) 10(1 0.456 ) ( ) ' s s s s G s + + + = ( ) 36.9 12.8 49.7 45 , '' '' = + = + = o o o o m c = '' h 解:由ess=1/kv=1/k≤ 0.1 , 有 K≥10; 10lg ( ) 6 ' '' a = −L c = a = 4 + − = , 1 1 arcsin a a m '' '' ( ) 180 90 c o o c = − − arctg = ' h